Chapter R Relativity
Conceptual Problems
3 • If event A occurs at a different location than event B in some reference frame, might it be possible for there to be a second reference frame in which they occur at the same location? If so, give an example. If not, explain why not. Determine the Concept Yes. Let the initial frame of reference be frame 1. In frame 1 let L be the distance between the events, let T be the time between the events, and let the +x direction be the direction of event B relative to event A. Next, calculate the value of L/T. If L/T is less than c, then consider the two events in a reference frame 2, a frame moving at speed v = L/T in the +x direction. In frame 2 both events occur at the same location. 5 • Two events are simultaneous in a frame in which they also occur at the same location. Are they simultaneous in all other reference frames? Determine the Concept Yes. If two events occur at the same time and place in one reference frame they occur at the same time and place in all reference frames. (Any pair of events that occur at the same time and at the same place in one reference frame are called a spacetime coincidence.) 11 •• Many nuclei of atoms are unstable; for example, 14C, an isotope of carbon, has a half-life of 5700 years. (By definition, the half-life is the time it takes for any given number of unstable particles to decay to half that number of particles.) This fact is used extensively for archeological and biological dating of old artifacts. Such unstable nuclei decay into several decay products, each with significant kinetic energy. Which of the following is true? (a) The mass of the unstable nucleus is larger than the sum of the masses of the decay products. (b) The mass of the unstable nucleus is smaller than the sum of the masses of the decay products. (c) The mass of the unstable nucleus is the same as the sum of the masses of the decay products. Explain your choice. Determine the Concept Because mass is converted into the kinetic energy of the fragments, the mass of the unstable nucleus is larger than the sum of the masses of the decay products. (a ) is correct.

Estimation and Approximation
13 •• In 1975, an airplane carrying an atomic clock flew back and forth at low altitude for 15 hours at an average speed of 140 m/s as part of a time-dilation experiment. The time on the clock was compared to the time on an atomic clock kept on the ground. What is the time difference between the atomic clock on the airplane and the atomic clock on the ground? (Ignore any effects that

213

214 Chapter R accelerations of the airplane have on the atomic clock that is on the airplane. Also assume that the airplane travels at constant speed.) Picture the Problem We can use the time dilation equation to relate the elapsed time in the frame of reference of the airborne clock to the elapsed time in the frame of reference of the clock kept on the ground. Use the time dilation equation to relate the elapsed time Δt according to the clock on the ground to the elapsed time Δt0 according to the airborne atomic clock: Because v

Substitute numerical values and evaluate v:

(2.998 ×10 v>

m/s (2.00 μs ) 1.50 km
8 2

)

= 1.20 × 108 m/s = 0.400c

218 Chapter R Event B can precede event A provided v > 0.400c.

Relativistic Energy and Momentum
41 • How much energy would be required to accelerate a particle of mass m from rest to (a) 0.500c, (b) 0.900c, and (c) 0.990c? Express your answers as multiples of the rest energy, mc2. Picture the Problem We can use Equation R-14 to find the energy required to accelerate this particle from rest to the given speeds.

From Equation R-14 we have:

K=

mc 2 1 − (v c )
2

− mc 2

⎛ ⎞ 1 =⎜ − 1⎟mc 2 2 ⎜ ⎟ ⎝ 1 − (v c ) ⎠

(a)Substitute numerical values and evaluate K(0.500c):

⎛ ⎞ 1 K (0.500c ) = ⎜ − 1⎟ mc 2 2 ⎜ ⎟ ⎝ 1 − (0.500c c ) ⎠ = 0.155mc 2

(b) Substitute numerical values and evaluate K(0.900c):

⎛ ⎞ 1 ⎜ K (0.900c ) = − 1⎟ mc 2 2 ⎜ ⎟ ⎝ 1 − (0.900c c ) ⎠ = 1.29mc 2

(c) Substitute numerical values and evaluate K(0.990c):

⎛ ⎞ 1 K (0.990c ) = ⎜ − 1⎟ mc 2 2 ⎜ ⎟ ⎝ 1 − (0.990c c ) ⎠ = 6.09mc 2

45 ••

(a) Show that the speed v of a particle of mass m and total energy E is

v ⎡ mc 2 given by = ⎢1 − c ⎢ E2 ⎣

(

)

2

⎤ ⎥ ⎥ ⎦

12

and that when E is much greater than mc2, this can

v mc 2 be approximated by ≈ 1 − . Find the speed of an electron with kinetic c 2E 2 energy of (b) 0.510 MeV and (c) 10.0 MeV.

(

)

2

Relativity 219
Picture the Problem We can solve the equation for the relativistic energy of a particle to obtain the first result and then use the binomial expansion subject to E >> mc2 to obtain the second result. In Parts (b) and (c) we can use the first expression obtained in (a), with E = E0 + K, to find the speeds of electrons with the given kinetic energies. See Table 39-1 for the rest energy of an electron.

(a) The relativistic energy of a particle is given by Equation R-15:

E=

mc 2 v2 1− 2 c

Solving for v/c yields:

2 ⎡ v mc 2 ⎤ = ⎢1 − ⎥ c E2 ⎥ ⎢ ⎣ ⎦

( )

12

(1)

Expand the radical expression binomially to obtain: v mc 2 = 1− c E2

(

)

2

1 mc 2 = 1− 2 E2

(

)

2

+ higher - order terms

Because the higher-order terms are much smaller than the 2nd-degree term when E >> mc2: (b) Solve equation (1) for v:

v mc 2 ≈ 1− c 2E 2

( )

2

v=c

(mc ) 1−
E2

2 2

Because E = E0 + K:

v = c 1−

E02 1 = c 1− 2 2 (E0 + K ) ⎛ K⎞ ⎜1 + ⎟ ⎜ E ⎟ 0 ⎠ ⎝
1 ⎛ 0.510 MeV ⎞ ⎜1 + ⎜ 0.511MeV ⎟ ⎟ ⎝ ⎠
2

For an electron whose kinetic energy is 0.510 MeV:

v(0.510 MeV ) = c 1 −

= 0.866c

220 Chapter R (c) For an electron whose kinetic energy is 10.0 MeV: v(10.0 MeV ) = c 1 − 1 ⎛ 10.0 MeV ⎞ ⎜1 + ⎜ 0.511MeV ⎟ ⎟ ⎝ ⎠
2

= 0.999c

General Problems
53 •• Particles called muons traveling at 0.99995c are detected at the surface of Earth. One of your fellow students claims that the muons might have originated from the Sun. Prove him wrong. (The proper mean lifetime of the muon is 2.20 μs.) Picture the Problem Your fellow student is thinking that the time dilation factor might allow muons to travel the 150,000,000,000 m from the Sun to Earth. You can discredit your classmate’s assertion by considering the mean lifetime of the muon from Earth’s reference frame. Doing so will demonstrate that the distance traveled during as many as 5 proper mean lifetimes is consistent with the origination of muons within Earth’s atmosphere.

The distance, in the Earth frame of reference, a muon can travel in n mean lifetimes τ is given by:

d =n

1 − (v c )

d0

2

=

nvτ 1 − (v c )
2

=

n(v c )cτ 1 − (v c )
2

Substitute numerical values for v, c, and τ and simplify to obtain:

d =n

(0.99995)(2.998 ×108 m/s)(2.20 μs ) = (66.0 km)n 2 1 − (0.99995) d = (66.0 km )(5) = 330 km , a distance

In 5 lifetimes a muon would travel a distance:

approximating a low-Earth orbit.

In 100 lifetimes, d ≈ 6600 km, or approximately one Earth radius. This relatively short distance should convince your classmate that the origin of the muons that are observed on Earth is within our atmosphere and that they certainly are not from the Sun.